ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-09-12

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Constructing an epsilon-NFA for $(a + b)^{*} a b b$

Example

After the standard Thompson construction and the obvious epsilon-edge simplifications, one convenient epsilon-NFA for $(a + b)^{*} a b b$ has states $q_{0}, q_{1}, q_{2}, q_{3}$ , start state $q_{0}$ , accepting state $q_{3}$ , and transitions

$q_{0} \overset{a}{\to} q_{0}, q_{0} \overset{b}{\to} q_{0}, q_{0} \overset{a}{\to} q_{1}, q_{1} \overset{b}{\to} q_{2}, q_{2} \overset{b}{\to} q_{3} .$

It is an epsilon-NFA because ordinary NFA's are the special case with no $ε$ -moves.

Facts & Assumptions

Given: The displayed machine with start state $q_{0}$ and accepting state $q_{3}$ .

[L1]

By Every regular expression has an equivalent epsilon-NFA, a regular expression has an equivalent epsilon-NFA.

Verification

technique · direct

1.1given

Any word of the form $x a b b$ with $x \in {a, b}^{*}$ is accepted: stay in $q_{0}$ while reading the prefix $x$ , then use the branch $q_{0} \overset{a}{\to} q_{1} \overset{b}{\to} q_{2} \overset{b}{\to} q_{3}$ on the final three letters.

2.1step 1.1

Conversely, every accepting run must finish with the forced chain $q_{0} \overset{a}{\to} q_{1} \overset{b}{\to} q_{2} \overset{b}{\to} q_{3}$ , because $q_{3}$ is reached only from $q_{2}$ by a $b$ -edge and $q_{2}$ is reached only from $q_{1}$ by a $b$ -edge. Therefore any accepted word ends in $a b b$ , and the earlier letters are exactly the ones read by the two loops at $q_{0}$ .

3.1L1step 1.1step 2.1∎

Hence this machine recognizes the words ending in $a b b$ , which is the language of $(a + b)^{*} a b b$ . This is the concrete instance promised by [L1].

Depends on

Every regular expression has an equivalent epsilon-NFA

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Jean Gallier and Jocelyn Quaintance, Introduction to the Theory of Computation: Some Notes for CIS511 (standard reference, not scraped)

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